Mathematical induction

4 August 2014One Comment

Monica Kochar

The market is flooded with learning material for primary and middle school maths lessons. However, for senior classes the question remains – what is the best design for the lessons? This article is an attempt to answer that question.

I have taken the topic of mathematical induction, for this is what students find difficult to grasp. ‘What exactly is happening here?’ is a question one hears often! The principle is taught followed by problem solving. Hardly anyone knows the reasoning!

Designing the lesson for senior classes
Step 1 – Modelling problem
Always begin a topic with a modelling problem. A modelling problem is based on real life and requires the use of maths in solving it. No formulas or formal terminology comes in at this point. Let the students use any method. Our first objective is to get them thinking.

Strategy
a. Divide students into groups of 4-5 (this ensures that no one is left behind).
b. Instruct each group to look out for its members and help where needed.
c. Hand out the following problem:
A post office sells stamps of 3p and 5p only. The minimum postage required in the city is 8p. The official proudly says “any postage above 8p can be paid using a combination of 3p and 5p stamps” or 3a+5b where a and b are whole numbers, 0 to ∞.
Check if he is right. Use any method to solve this.

d. Spend about 5 minutes discussing the problem. If needed, give a small hint.Check for a few numbers:
8=3+5 (one of 3 and one of 5) or 3a+5b
9=3+3+3 (three of 3 and none of 5) or 3a+5b
10=5+5 (none of 3 and two of 5) or 3a+5b
e. Set your timer to 20 minutes so that students stay focussed. If possible, display the timer where everyone can see it or ask a student in each group to be the time keeper.
f. After 20 minutes, ask each group how far they went. If a group solved the problem, then let them share it with the whole class.
g. If no one could solve it, teacher could give more time or move on with hints and solution.

(Hints for the teacher)
After checking for 8, 9, and 10, check if the next cluster also shows the same combination
11=3+3+5=3a+5b
12=3×4=3a+5b
13=3+5+5=3a+5b
It does!
Check for the next cluster of 3 numbers 14, 15, 16…it will work!

Step 2 – Reasoning
Strategy
a. Teacher to explain to the students why this works.
b. Let students be seated in groups
c. At the end of the teacher’s explanation, ask each group to explain the solution to each other ensuring that everyone has understood it.

Teacher explanation
8, 9, 10…is a combination of 3 and 5 and hence = 3a+5b
8=3a+5b, as shown above
11=8+3 and hence = 3a+5b + 3 = 3(a+1) + 5b which is in the form 3a+5b again
Similarly 12 is 3 up 9 and hence would be in the form 3a+5b again
In fact 11, 12, 13…14, 15, 16…every cluster is 3 up the previous cluster. Since the first cluster is a part of 3a+5b, every subsequent cluster is also a part of 3a+5b
Hence, it is true that every postage above 8 can be paid by a combination of 3 and 5 postage stamps.

Step 3 – Students independently try another problem
Let students try a new problem, including reasoning, to get the idea.
Hand out the following problem:
Suppose an ATM machine has only Rs. 2 and Rs. 5 bills. You can type in the amount you want, and the machine will figure out how many twos and fives it needs to process your request. The ATM machine can generate any output amount n ≥ 4.

Let the students brainstorm, solve and share with the class. As before, set the timer to a stipulated time.

Step 4 – Generalization
This is where algebraic generalization steps in, building the way to the principle.
Take the post office problem. It says that a combination of 3 and 5 stamps is equal to 3a + 5b where a and b are whole numbers 0 to ∞. But how can we say that this is true for any cluster without trying it out for each number? This is where maths helps!

Let us see…
This is true for the first number 8 = 3(a=1) + 5(b=1)
Let us assume for any natural number k, k=3a+5b. k>8
Would it true that the next number k+1 is also a combination of 3 and 5?
K=3a+5b
K+1=3a+5b+1 OR 3a + 6+5(b-1) OR 3(a+2) + 5(b-1)
OR
3a + 5b for any numbers a and b!
Now 8=3a+5b
If k=3a+5b, then the next number k+1 is also 3a+5b
So any number above 8 = 3a+5b.

Step 5 – Mathematical induction
This is where the teacher introduces the principle of mathematical induction.
This is what we have done!
1. Check if the rule is true for the first possible number, ‘a’. This is called P(a)
2. Show that if it is true for ‘k’, a random number, it is also true for ‘k+1’, the next number.

Hence it is true for all numbers starting from ‘a’. This is called inductive logic, where one starts with some examples and builds a generalization. (As against deductive logic where one starts with generalization and then checks it using some examples.)

Example
Prove that 1+2+3+…+n = n(n+1)/2 for any natural number n
1. Check if true for n=1. P(1) is true
2. Show that if it is true for ‘k’, a random number, it is also true for ‘k+1’, the next number.
If it is true for 1 and true for consecutive numbers, then it is true for 2 and hence for 3 and 4 … this is what we have shown!
This is what mathematicians call the Principle of Mathematical Induction.

Note to teachers
You will be doing a lot of good to the students’ understanding of maths using this method for you will be connecting maths to real life. By working in groups, social development happens. Thinking out a problem develops reasoning skills. Granted that in senior school time is a constraint, however spending two lessons on such a thinking level is bound to increase students’ enthusiasm and hence increase the pace of work!
Try it once!

The author has been a teacher for 19 years and is on a sabbatical from ‘schooling’. She is now a freelance math curriculum developer and is experimenting with strategies to make math accessible to all. She can be reached at reachmonica@gmail.com and has a blog called http://humanemaths.blogspot.in.